# The Kernel of a Homomorphism

" The kernel is important because it controls the entire homomorphism. It tells us not only which elements of G are mapped to the identity in G', but also which pairs of elements have the same image in G'. " (Taken from Algebra by Artin.)

How does the kernel tell us which pairs of elements have the same image in G'?

What I tried:

let $$f$$ be a homomorphism such that $$f(a) = f(b) = c$$. $$c$$ is in $$G'$$, so it has an inverse $$c^{-1}$$. So then, $$c^{-1}f(a) = c^{-1}f(b) = cc^{-1} = 1$$.

But, this is tedious to compute, and I didn't use the kernel to do it.

• $f(a)=f(b)=c\implies f(ab^{-1})=e'$. – lulu Jun 9 '19 at 18:51

The point is that $$f(a)=f(b)$$ if and only if $$ab^{-1}\in \ker f$$. So, if you have checked beforehand that $$\ker f=\{e\}$$, then $$f(a)=f(b)$$ automatically implies that $$a=b$$, which means that $$f$$ is injective. And it works backwards too, if $$f$$ is injective, then $$\ker f=\{e\}$$, because $$f(a)=e =f(e)$$ implies $$a=e$$.
If $$c$$ belongs to the image, then there is a $$g_0\in G$$ such that $$f(g_0)=c$$. But then$$\{g\in G\mid f(g)=c\}=\{g_0h\mid h\in\ker f\}.$$
• does this imply that $g = g_0h$? (for every $h \; \epsilon$ ker $f$ ?) – Jess Jun 9 '19 at 18:58
• What is $g$? What is $h$? – José Carlos Santos Jun 9 '19 at 18:59
• $g$ is every element of $G$ such that $f(g) = c$. $h$ is any element of $G$ in the Kernel? – Jess Jun 9 '19 at 19:01
• What I am saying is that an element $g$ of $G$ has the property that $f(g)=c$ when and only when $g$ can be written as $g_0h$, where $h$ is an element of $\ker f$. – José Carlos Santos Jun 9 '19 at 19:03